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Let x and y be nonnegative real numbers. If x^2+3y^2=18, then find the maximum value of xy.

 Feb 16, 2024

Best Answer 

 #2
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As we need to find the max value of a product we can use AM-GM Inequality. the inequality states that for any real numbers \(x_1, x_2, \ldots, x_n \geq 0\),
\(\LARGE\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}\)

with equality if and only if \(x_1 = x_2 = \cdots = x_n\)

Thus using this we write:
\(\Large\frac{x^2+3y^2}{2}\ge\sqrt[2]{x^2\cdot3y^2}\)

\(\)Plugging in values and solving-->
\(\sqrt3xy\le9\\ \boxed{xy\le3\sqrt3}\)

 Feb 16, 2024
 #1
avatar+53 
+1

I guess draw the ellipse

 Feb 16, 2024
 #2
avatar
+2
Best Answer

 

As we need to find the max value of a product we can use AM-GM Inequality. the inequality states that for any real numbers \(x_1, x_2, \ldots, x_n \geq 0\),
\(\LARGE\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}\)

with equality if and only if \(x_1 = x_2 = \cdots = x_n\)

Thus using this we write:
\(\Large\frac{x^2+3y^2}{2}\ge\sqrt[2]{x^2\cdot3y^2}\)

\(\)Plugging in values and solving-->
\(\sqrt3xy\le9\\ \boxed{xy\le3\sqrt3}\)

EnormousBighead Feb 16, 2024

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